A newspaper conducted a statewide survey concerning a proposal to raise taxes in order to prevent budget cuts to education

Question

A newspaper conducted a statewide survey concerning a proposal to raise taxes in order to prevent budget cuts to education. The newspaper took a random sample (assume it is an SRS) of 500 registered voters and found that 155 would vote to raise taxes. Let p represent the proportion of registered voters in the state that would vote to raise taxes. (Round your answers to 4 decimal places, if needed.)

(a) The standard error for the sample proportion is .

(b) The margin of error for an 87% confidence interval for p is .

Answer 1

Solution :

a)

Standard error of the sample proportion is given as follows :

S E p^ = √[(p^ )(1- p ^)]/[n]

Where p^ is the sample proportion and n is sample size

Sample proportion of voters who would vote to raise taxes is given by

p^ = 155/500 = 0.31

n=900

Hence the standard error for the sample proportion is given by

S E p^ = √[( 0.31 )(1- 0.31)] /900

= √ 0.0002376666

=0.015416

=0.01542

Standard error for the sample is 0.01542

S.E = S / √n

= 500/ √ 150

= 500/√ 150

Standard error = standard deviation / √ sample

= 5/√n

Assuming 5 = 500 = 500/√150

= 40.82

Mergin of error (87%) = x— +_ z s/√n

Z score = 1.5141025

MoE = z * √[p*(1-p )] / [√n ]

Z= 1.514 Alpha = 87% n= 500

1.514 × √[ 0.13 × (1-0.13)] / √ [500]

= 0.0227

b)

The mergin error for an 87% for confident interval for p is given by

E = Z 0.13/2 √ [p^ (1-p^ )] / [ n ]

Where Z 0.13/2 is critical value to construct 87% confidence interval using z-table we get Z (0.13/2) = 1.5141

Margin of error is

E = 1.5141 × √ [0.31 (1-0.31)] / 900

= 0.02376

The Margin error for an 87% confidence interval for p is 0.02376